# Breaking of diffeomorphism invariance after fixing a background metric

The Lagrangian for the gravitational field in absence of matter is the following $$L=1/k\int dx^4 \sqrt g R,$$ where $k=\sqrt G$, $g$ is the determinant of the metric and $R$ the Ricci scalar. It's possible to fix a background metric like $\eta_{uv}$ and then study the perturbations $h_{uv}$ around it by $$g_{uv}=\eta_{uv}+kh_{uv}$$ The Lagrangian becomes $$L=L^{0}+kL^{1}+k^{2}L^{2}+.......$$ which can be interpreted as an effective field theory of self-interacting particles called gravitons. Now, given the transformation law of $h_{uv}$, how is it possible to say the entire Lagrangian is invariant, order by order, under local diffeomorphisms? Of course the symmetry is still there, but I was wondering if there is some kind of Spontaneus Symmetry Breaking associated with the perturbation field $h_{uv}$ and the diffeomorphisms group. The procedure resemble the SSB for the Higgs Boson, where the Lagrangian is $$L=\partial_{u}\phi\partial^{u}\phi - m^{2}\phi^{2}+\lambda\phi^{4}$$ This Lagrangian is invariant under parity in $\phi$, but after the redefinition around the vacuum $v$, the minimum of the potential, you deal with $\phi=v+\delta\phi$ and the Lagrangian in δϕ is no more parity invariant. Does this happen in the previous example after fixing a background?

• See Schwartz's Quantum Field Theory and the Standard Model, section 8.7.2. (and maybe section 22.4) – AccidentalFourierTransform Jan 3 '17 at 17:42
• I have received only a partial answer to my previous question, so I tried to explain myself in a better way. – Yildiz Jan 3 '17 at 17:51
• Ok, I' ll try to look at Schwartz. Anyway, It seems to me that fixing a background metric breaks diffeomorphism invariance of the Lagrangian. – Yildiz Jan 3 '17 at 17:55
• @Yildiz you cannot lose symmetries by field redefinitions. In this case, the redefinition $g=\eta+h$ is linear; this makes the symmetry less obvious/manifest but it is there. – AccidentalFourierTransform Jan 3 '17 at 18:41
• This post has been discussed on Meta, see here. – AccidentalFourierTransform Jan 5 '17 at 17:33

No, there is no SSB after a field redefinition. Recall that SSB is a dynamical effect. You cannot trigger dynamical effects by changing your coordinates. In the scalar case the SSB is triggered by a quadratic term with "the wrong sign", not by the change of variables $\phi=v+\delta\phi$. The new physics are more transparent in the new coordinates; but the SSB is not caused by changing into the new coordinates: the symmetry breaks whether you define $\phi=v+\delta\phi$ or not.

In other words, physics are independent of coordinates. Using $g\to\eta+h$ leaves the physics invariant. You can also change $g\to g_0+g_1$, and choose any background $g_0$. The flat background is convenient, but in the literature you'll also find people that consider more general backgrounds; for example, $g_0$ can be taken to be the metric of an asymptotically flat space-time. In any case, the dynamics are determined by the Lagrangian, not by the coordinates.

• I understand what you are saying, anyway you will agree with me that when I quantize a system in order to deal with particles I have to work around a minimum of the potential, and while using $\delta\phi$ I have to be careful because is not true that the Lagrangian is parity invariant when "reads" in the perturbations: of course the initial symmetry is still there, but is less manifest like in the case of the Higgs Boson. I was wondering if the same happens in my case – Yildiz Jan 11 '17 at 14:19
• Yes, I agree with you: when written in terms of $g_{\mu\nu}$ the gauge symmetry is obvious, but when written in terms of $h_{\mu\nu}$ the symmetry is no longer obvious. The symmetry is not manifest but it is there. The analysis of the diff. inv. of the Einstein-Hilbert Lagrangian is analysed in detail in Schwartz's book. See also General Covariance and Background Independence in Quantum Gravity, by M. Bärenz. – AccidentalFourierTransform Jan 11 '17 at 14:39
• I have taken a look at Barenz and Schwartz, but I didn't find an answer. My main problem is the following: the transformation law $h_{uv}^{'}=h_{uv}+\partial_{u}\xi_{v}-\partial_{v}\xi_{u}$ is it the diffeomorphism symmetry applied naively to $h_{uv}$, or is it the hidden symmetry, the correct one, which preserves the diff. invariance for $g_{uv}$? – Yildiz Jan 12 '17 at 12:58
• In the Higgs Boson example is the same as transforming $\delta\phi$ in $-\delta\phi$ or $-\delta\phi-2v$: the first is parity applied naively to perturbations which is violated, the second one is the hidden symmetry which ensures that parity is preserved for the Lagrangian in $\phi$. It could appear a game of words, but it's not, please let me know if you understand me and I'll try to be clearer. – Yildiz Jan 12 '17 at 12:59
• For sure the transformation law that I wrote for $h_{uv}$ is not the hidden one, because it's only the linearization of the true hidden one $h_{uv}^{'}=g_{uv}^{'}-\eta_{uv}$ where $g_{uv}$ transforms in the usual way. If so it seems that diffeomorphism invariance is violated. – Yildiz Jan 12 '17 at 13:06

Schematically, the Einstein-Hilbert action is given by,

$$S \sim \int d^D x \, \sqrt{|\det g_{\mu\nu}|} \, \mathcal R$$

for a metric, $g_{\mu\nu}$. Now, as noted in previous questions and by the OP, one can expand the field as,

$$g_{\mu\nu} = \eta_{\mu\nu} + h_{\mu\nu}$$

and since the inverse metric is an infinite series in $h_{\mu\nu}$, one obtains an infinite number of terms in the Einstein-Hilbert action expressed in this form. This procedure does not violate diffeomorphism invariance, as it is a mere field redefinition and we know all the terms sum to give $S$.

We can express any metric in the form $\eta_{\mu\nu} + h_{\mu\nu}$, it is as trivial as expressing a scalar $\phi$ in terms of two scalars, one of which we can choose completely freely, e.g. $\phi = \eta + h$.

• I agree that a symmetry remains after field redefinition, anyway it could be less manifest. The same happen with Spontaneuos Symmetry Breaking for the Higgs Boson: the Lagrangian has $\phi$ as field and it's invariant under parity, but after the redefinition $\phi=v+\delta\phi$, the Lagrangian in $\delta\phi$ is not parity invariant: of course the symmetry still exist but is hidden. Does this happen in our case? Not only this: in the Higgs Boson example $v$ was the minimum of a potential, but here we don't have a potential, so why choosing $\eta_{uv}$ and not another metric? – Yildiz Jan 6 '17 at 14:21
• Of course, $\eta_{uv}$ is a natural choice as a background, anyway there is no dynamical procedure which gives it as a background. It's not derived as minimum from a potential in our Lagrangian unlike the vacuum $v$ for the Higgs Boson and other fields. – Yildiz Jan 6 '17 at 14:34
• @Yildiz It is simply convenient to expand around $\eta_{\mu\nu}$. In gravitational perturbation theory, we actually expand around a general metric, $g_{\mu\nu} = {g_{0}}_{\mu\nu} + h_{\mu\nu}$. Things simplify for ${g_{0}}_{\mu\nu}=\eta_{\mu\nu}$. – JamalS Jan 6 '17 at 14:44
• For sure it's convenient in many situations, but usually this field redefinition is accompanied in many books (e.g. Schwartz) with the claim that GR is not renormalizable, and that in order to resolve this problem we have to find a UV completion. – Yildiz Jan 6 '17 at 14:58
• @Yildiz The division between string theory and LQG has nothing to do with that, because string theory makes no attempt to quantise a theory of $g_{\mu\nu}$; Einstein gravity emerges as the renormalisation group flow of the string in a non-linear sigma model. – JamalS Jan 6 '17 at 15:02